Sears, Zemansky, Young & Freedman.

350 Pages·2009·5.06 MB·English

Física III – Sears, Zemansky,
Young & Freedman.
PHYSICS ACT.

http//physicsact.wordpress.com

Capítulo 22

22.1: a)
C.
Nm
75
.1
60
cos
)
m
(0.250

N/C)
14
(
2
2
=
°
=
⋅
=
Φ
A
E

r
r

b) As long as the sheet is flat, its shape does not matter.
ci) The maximum flux occurs at an angle
°
= 0
φ
between the normal and field.
cii) The minimum flux occurs at an angle
°
= 90
φ
between the normal and field.
In part i), the paper is oriented to “capture” the most field lines whereas in ii) the
area is oriented so that it “captures” no field lines.

22.2: a)
n
A
A
E
ˆ
where
cos
A
θ
EA
=
=
⋅
=
Φ

r
r
r

C
m
N
32
9.
36
cos
m)
(0.1
)
C
N
10
4
(
(back)
ˆ
ˆ

C
m
N
32
9.
36
cos
m)
(0.1
)
C
N
10
4
(
front)
(
ˆ
ˆ

0
90
cos
m)
(0.1
)
C
N
10
4
(
bottom)
(
ˆ
ˆ

C
m
N
24
)
9.
36
(90
cos
m)
(0.1
)
C
N
10
4
(
right)
(
ˆ
ˆ

0
90

cos
m)
(0.1
)
C
N
10
4
(
top)
(
ˆ
ˆ

C
m
N
24
)
9
36
90
(
cos
m)
(0.1
)
C
N
10
4
(
left)
(
ˆ
ˆ

2
2
3

2
2
3

2
3

2
2
3

2
3
S

2
2
3

6
6

5
5

4
4

3
3

2
2

1
1

⋅
−
=
°
×
−
=
Φ
−
=

⋅
=
°
×
+
=
Φ
+
=

=
°
×
=
Φ
−
=

⋅
+
=
°
−
°
×
+
=
Φ
+
=

=
°
×
−
=
Φ
+
=

⋅
−
=
°
−
×
−
=
Φ
−
=

S
S

S
S

S
S

S
S

S

S
S
.

i
n

i
n

k
n

j
n

k
n

j
n

b) The total flux through the cube must be zero; any flux entering the cube must also
leave it.

22.3: a) Given that
length
edge
,
,ˆ
D
ˆ
C
ˆ
B
A
E
k
j
i
E

r
r
r

⋅
=
Φ
−
+
−
=
L, and

.
BL
ˆ
ˆ
ˆ

.
BL
ˆ
ˆ
ˆ

.
DL
ˆ
ˆ
ˆ

.
CL
ˆ
ˆ
ˆ

.
DL
ˆ
ˆ
ˆ

.
CL
ˆ
ˆ
ˆ

2
6

2
5

2
4

2
3

2
2

2
1

6
6

5
5

4
4

3
3

2
2

1
1

+
=
⋅
=
Φ
⇒
−
=

−
=
⋅
=
Φ
⇒
+
=

+
=
⋅
=
Φ
⇒
−
=

+
=
⋅
=
Φ
⇒
+
=

−
=
⋅
=
Φ
⇒
+
=

−
=
⋅
=
Φ
⇒
−
=

S
S

S
S

S
S

S
S

S
S

S
S

A

A

A

A

A

A

n
E
i
n

n
E
i
n

n
E
k
n

n
E
j
n

n
E
k
n

n
E
j
n

r

r

r

r

r

r

b) Total flux ∑ =
=
Φ
=
6

1
i
0
i

22.4:
C.
Nm
16
.6
70
cos
)
m
(0.240
)
C
N
0.
75
(
2
2
=
°
=
⋅
=
Φ
A
E

r
r

22.5: a)
C.
Nm
10
71
.2
)
2
(
2
5
m)
(0.400
C/m)
10
00
.6
(
2
0

6

0
0
×
=
=
=
=
⋅
=
Φ

−
×
λ
λ
ε
ε
l
r
πε
πrl
A
E

r
r

b) We would get the same flux as in (a) if the cylinder’s radius was made larger—the
field lines must still pass through the surface.
c) If the length was increased to
m,
800
.0
=
l
the flux would increase by a factor of
two:
C.
Nm
10
5.42
2
5
×
=
Φ

22.6: a)
C.
Nm
452
C)
10
00
.4
(
2
0
9
0
1
1
=
×
=
=
Φ
−
ε
ε
q
S

b)
C.
Nm
881
C)
10
80
.7
(
2
0
9
0
2
2
−
=
×
−
=
=
Φ
−
ε
ε
q
S

c)
C.
Nm
429
C)
10
)
80
.7
00
.4
((
)
(
2
0
9
0
2
1
3
−
=
×
−
=
+
=
Φ
−
ε
ε
q
q
S

d)
C.
Nm
723
C)
10
)
40
.2
00
.4
((
)
(
2
0
9
0
2
1
4
=
×
+
=
+
=
Φ
−
ε
ε
q
q
S

e)
C.
Nm
158
C)
10
)
40
.2
80
.7
00
.4
((
)
(
2
0
9
0
3
2
1
5
−
=
×
+
−
=
+
+
=
Φ
−
ε
ε
q
q
q
S

f) All that matters for Gauss’s law is the total amount of charge enclosed by the
surface, not its distribution within the surface.

22.7: a)
C.
Nm
10
07
.4
C)
10
60
.3
(
2
5
0
6
0
×
−
=
×
−
=
=
Φ
−
ε
ε
q

b)
C.
10
6.90
)
C
Nm
780
(
9
2
0
0
0
−
×
=
=
Φ
=
⇒
=
Φ
ε
ε
q
ε
q

c)
No. All that matters is the total charge enclosed by the cube, not the details of
where the charge is located.

22.8: a) No charge enclosed so
0
=
Φ

b)
C.
Nm
678
Nm
C
10
8.85
C
10
00
.6
2
2
2
12

9

0

2
−
=
×
×
−
=
=
Φ
−

−

ε
q

c)
C.
Nm
226
Nm
C
10
8.85
C
10
)
00
.6
00
.4
(
2
2
2
12

9

0

2
1
−
=
×
×
−
=
+
=
Φ
−

−

ε
q
q

22.9: a) Since E

r

is uniform, the flux through a closed surface must be zero. That is:

∫
=
∫
⇒
=
∫
=
=
⋅
=
Φ
.0
0
0
0
1
ρdV
ρdV
d
ε
ε
q
A
E

r
r

But because we can choose any volume we

want, ρ must be zero if the integral equals zero.
b) If there is no charge in a region of space, that does NOT mean that the electric field
is uniform. Consider a closed volume close to, but not including, a point charge. The field
diverges there, but there is no charge in that region.

22.10: a) If
0
>
ρ
and uniform, then q inside any closed surface is greater than zero.

∫
>
⋅
⇒
>
Φ
⇒
0
0
A
E

r
r

d
and so the electric field cannot be uniform, i.e., since an

arbitrary surface of our choice encloses a non-zero amount of charge, E must depend on
position.
b) However, inside a small bubble of zero density within the material with density ρ ,
the field CAN be uniform. All that is important is that there be zero flux through the
surface of the bubble (since it encloses no charge). (See Exercise 22.61.)

22.11:
C.
Nm
10
08
.1
C)
10
60
.9
(
2
6
0
6
0
sides
6
×
=
×
=
=
Φ
−
ε
ε
q
But the box is

symmetrical, so for one side, the flux is:
.
C
Nm
10
80
.1
2
5
side
1
×
=
Φ

b) No change. Charge enclosed is the same.

22.12: Since the cube is empty, there is no net charge enclosed in it. The net flux,
according to Gauss’s law, must be zero.

22.13:
0
encl ε
Q
E =
Φ

The flux through the sphere depends only on the charge within the sphere.

nC
3.19
)
C
m
N
360
(
2
0
0
encl
=
⋅
=
Φ
=
ε
ε
Q
E

22.14: a)
.
C
N
44
.7
m)
(0.550
C)
10
50
.2
(
4
1
4
1
m)
0.1
m
450
.0
(
2

10

0
2
0
=
×
=
=
+
=

−

πε
r
q
πε
r
E

b)
0
=
E
r

inside of a conductor or else free charges would move under the influence of

forces, violating our electrostatic assumptions (i.e., that charges aren’t moving).

22.15: a)
m.
62
.1
C
N
614
C)
10
180
.0
(
4
1
4
1
|
|
4
1
|
|

6

0
0
2
0
=
×
=
=
⇒
=

−

πε
E
q
πε
r
r
q
πε
E

b) As long as we are outside the sphere, the charge enclosed is constant and the sphere
acts like a point charge.

22.16: a)
C.
10
56
.7
)
m
(0.0610
)
C
N
10
40
.1(
/
8
2
5
0
0
0
−
×
=
×
=
=
⇒
=
=
Φ
ε
EA
ε
q
ε
q
EA

b) Double the surface area:
C.
10
51
.1
)
m
(0.122
)
C
N
10
40
.1(
7
2
5
0
−
×
=
×
= ε
q

22.17:
C.
10
27
.3
m)
(0.160
)
C
N
1150
(
4
4
9
2
0
2
0
4
1
2
0

−
×
=
=
=
⇒
=
πε
Er
πε
q
E
r

q
πε
So the

number of electrons is:
.
10
04
.2
10

C
10
60
.1

C
10
27
.3
e
19

9
×
=
=
−

−

×

×
n

22.18: Draw a cylindrical Gaussian surface with the line of charge as its axis. The
cylinder has radius 0.400 m and is 0.0200 m long. The electric field is then 840 N/C at
every point on the cylindrical surface and directed perpendicular to the surface. Thus

∫
=
=
⋅
)
2
)(
(
)
)(
(
cylinder
πrL
E
A
E
d s
E
r
r

/C
m
N
42.2
m)
(0.0200
m)
(0.400
)
(2
N/C)
840
(
2
⋅
=
=
π

The field is parallel to the end caps of the cylinder, so for them
0
=
⋅
∫
s
E
r
r

d
. From

Gauss’s law:

C
10
74
.3

)
C
m
N
2.
42
(
)
m
N
C
10
854
.8
(

10

2

2

2
12
0

−

−

×
=

⋅
⋅
×
=
Φ
=
E
ε
q

22.19:

r
E
λ
2
1

0
πε
=

22.20: a) For points outside a uniform spherical charge distribution, all the charge can
be considered to be concentrated at the center of the sphere. The field outside the sphere
is thus inversely proportional to the square of the distance from the center. In this case:

C
N
53
cm
0.600
cm
0.200
)
C
N
480
(

2
=








=
E

b) For points outside a long cylindrically symmetrical charge distribution, the field is
identical to that of a long line of charge:

,
2
λ

0r
πε
E =

that is, inversely proportional to the distance from the axis of the cylinder. In this case

C
N
160
cm
0.600
cm
0.200
)
C
N
480
(
=








=
E

c) The field of an infinite sheet of charge is
;
2
/
0ε
σ
E =
i.e., it is independent of the
distance from the sheet. Thus in this case
.
C
N
480
=
E

22.21: Outside each sphere the electric field is the same as if all the charge of the sphere
were at its center, and the point where we are to calculate E

r

is outside both spheres.

2
1 and E
E

r
r

are both toward the sphere with negative charge.

sphere.

charged

negatively

the
toward
,
C
N
10
06
.8

C
N
10
471
.5
m)
(0.250
C
10
80
.3
|
|

C
N
10
591
.2
m)
(0.250
C
10
80
.1
|
|

5
2
1

5
2

6

2
2

2
2

5
2

6

2
1

1
1

×
=
+
=

×
=
×
=
=

×
=
×
=
=

−

−

E
E
E

k
r
q
k
E

k
r
q
k
E

22.22: For points outside the sphere, the field is identical to that of a point charge of the
same total magnitude located at the center of the sphere. The total charge is given by
charge density × volume:

C
10
60
.1
m)
150
.0
)(
3
4
)(
m
C
n
50
.7
(
10
3
3
−
×
=
=
π
q

a) The field just outside the sphere is

C
N
4.
42
m)
(0.150
C)
10
(1.06
)
/C
m
N
10
9
(
4
2

10
2
2
9

2
0
=
×
⋅
×
=
=

−

r
πε
q
E

b) At a distance of 0.300 m from the center (double the sphere’s radius) the field will
be 1/4 as strong: 10.6
C
N

c) Inside the sphere, only the charge inside the radius in question affects the field. In
this case, since the radius is half the sphere’s radius, 1/8 of the total charge contributes to
the field:

C
N
2.
21
m)
(0.075
C)
10
06
.1(
)
8
/
1(
)
C
/
m
N
10
9
(

2

10
2
2
9
=
×
⋅
×
=

−
E

22.23: The point is inside the sphere, so
3
/ R
kQr
E =
(Example 22.9)

nC
2.
10
)
m
100
.0
(
m)
(0.220
)
C
N
950
(
3
3
=
=
=
k
kr
ER
Q

22.24: a) Positive charge is attracted to the inner surface of the conductor by the charge
in the cavity. Its magnitude is the same as the cavity charge:
nC,
00
.6
inner
+
=
q
since

0
=
E
inside a conductor.
b) On the outer surface the charge is a combination of the net charge on the conductor
and the charge “left behind” when the
nC
00
.6
+
moved to the inner surface:
nC.
1.00
nC
6.00
nC
00
.5
inner
tot
outer
outer
inner
tot
−
=
−
=
−
=
⇒
+
=
q
q
q
q
q
q

22.25:
3
2
S
and
S
enclose no charge, so the flux is zero, and electric field outside the

plates is zero. For between the plates,
1
S shows that:
.
0
0
0
ε
σ
E
ε
A
σ
ε
q
EA
=
⇒
=
=

22.26: a) At a distance of 0.1 mm from the center, the sheet appears “infinite,” so:

∫
=
×
=
=
⇒
=
=
⋅

−
.
C
N
662
m)
800
.0
(
2
C
10
50
.7
2
2
2
0

9

0
0
ε
A
ε
q
E
ε
q
A
E
dA
E
r
r

b) At a distance of 100 m from the center, the sheet looks like a point, so:

.
C
N
10
75
.6
m)
(100
C)
10
50
.7
(
4
1
4
1
3
2

9

0
2
0

−
−
×
=
×
=
≈
πε
r
q
πε
E

c) There would be no difference if the sheet was a conductor. The charge would
automatically spread out evenly over both faces, giving it half the charge density on any
as the insulator

0
0
2
:
:).
(
ε
σ
ε
σ
c
E
σ
=
=
near one face. Unlike a conductor, the insulator is the

charge density in some sense. Thus one shouldn’t think of the charge as “spreading over
each face” for an insulator. Far away, they both look like points with the same charge.

22.27: a)
.
2
2
λ
πR
σ
L
Q

πRL
Q

A
Q
σ
=
=
⇒
=
=

b) ∫
=
⇒
=
=
=
⋅
.
2
)
2
(

0
0
0
rε
σR
E
ε
πRL
σ
ε
Q
πrL
E
d A
E

c) But from (a),
,
so
,
2
λ
0
2
λ
r
πε
E
R
=
=
π
σ
same as an infinite line of charge.

22.28: All the
s'σ
are absolute values.
(a) at

0

1

0

4

0

3

0

2
2
2
2
2
:
ε
σ
ε
σ
ε
σ
ε
σ
A
E
A
−
+
+
=

left.

the
to
C
N
10
82
.2

)
m
C
6
m
C
4
m
C
2
m
C
5
(
2
1

)
(
2
1

5

2
2
2
2

0

1
4
3
2
0

×
=

−
+
+
=

−
+
+
=

µ
µ
µ
µ
ε

σ
σ
σ
σ
ε
EA

(b)

left.

the
to
C
N
10
95
.3

)
m
C
5
m
C
4
m
C
2
m
C
6
(
2
1

)
(
2
1
2
2
2
2

5

2
2
2
2

0

2
4
3
1
0
0

2

0

4

0

3

0

1

×
=

−
+
+
=

−
+
+
=
−
+
+
=

µ
µ
µ
µ
ε

σ
σ
σ
σ
ε
ε
σ
ε
σ
ε
σ
ε
σ
EB

(c)

left

the
to
C
N
10
69
.1

)
m
C
6
m
C
4
m
C
2
m
C
5
(
2
1

)
(
2
1
2
2
2
2

5

2
2
2
2

0

1
4
3
2
0
0

1

0

4

0

3

0

2

×
=

−
−
+
=

−
−
+
=
−
−
+
=

µ
µ
µ
µ
ε

σ
σ
σ
σ
ε
ε
σ
ε
σ
ε
σ
ε
σ
EC

22.29: a) Gauss’s law says +Q on inner surface, so
0
=
E
inside metal.
b) The outside surface of the sphere is grounded, so no excess charge.
c) Consider a Gaussian sphere with the –Q charge at its center and radius less than
the inner radius of the metal. This sphere encloses net charge –Q so there is an electric
field flux through it; there is electric field in the cavity.

d) In an electrostatic situation
0
=
E
inside a conductor. A Gaussian sphere with
the
Q
−
charge at its center and radius greater than the outer radius of the metal encloses
zero net charge (the
Q
−
charge and the
Q
+
on the inner surface of the metal) so there is
no flux through it and
0
=
E
outside the metal.

e) No,
0
=
E
there. Yes, the charge has been shielded by the grounded
conductor. There is nothing like positive and negative mass (the gravity force is always
attractive), so this cannot be done for gravity.

22.30: Given
,ˆ
)
m
)
C
N
(
00
.3
(
ˆ
)
m
)
C
N
(
00
.5
(
k
i
E
z
x
⋅
+
⋅
−
=

r

edge length

,
m
300
.0
=
L
,
m
300
.0
=
L
and
.0
ˆ
ˆ
ˆ
1
1
1
=
⋅
=
Φ
⇒
−
=
A
S
s
n
E
j
n

r

⋅
=
⋅
=
Φ
⇒
+
=
)
C
N
(
00
.3
(
ˆ
ˆ
ˆ
2
1
2
A
S
S
n
E
k
n

r

=
=
z
z
)
m
)
C
N
(
27
.0
(
)
m
300
.0
)(
m
2

.
m
)
C
N
(
081
.0
)
m
300
.0
)(
m
)
C
N
(
27
.0
(
2
=

.0
ˆ
ˆ
ˆ
3
3
3
=
⋅
=
Φ
⇒
+
=
A
S
S
n
E
j
n

r

).
0
(
0
)
m
)
C
N
(
27
.0
(
ˆ
ˆ
ˆ
4
4
4
=
=
⋅
−
=
⋅
=
Φ
⇒
−
=
z
z
A
S
S
n
E
k
n

r

x
x
A
S
S
)
m
)
N/C
(
45
.0
(
)
m
300
.0
)(
m
)
C
N
(
00
.5
(
ˆ
ˆ
ˆ
2
5
5
5
⋅
−
=
⋅
−
=
⋅
=
Φ
⇒
+
=
n
E
i
n

r

).
m
)
C
N
(
135
.0
(
)
m
300
.0
)(
m
)
N/C
(
45
.0
(
2
⋅
−
=
⋅
−
=

).
0
(
0
)
m
)
C
N
(
45
.0
(
ˆ
ˆ
ˆ
6
6
6
=
=
⋅
+
=
⋅
=
Φ
⇒
−
=
x
x
A
S
S
n
E
i
n

r

b) Total flux:

C
Nm
054
.0
m
)
C
N
(
)
135
.0
081
.0
(
2
2
5
2
−
=
⋅
−
=
Φ
+
Φ
=
Φ

C
10
78
.4
13
0
−
×
−
=
Φ
= ε
q

22.31: a)

b) Imagine a charge q at the center of a cube of edge length 2L. Then:
.
/
0ε
q
=
Φ

Here the square is one 24th of the surface area of the imaginary cube, so it intercepts 1/24
of the flux. That is,
.
24
0
ε
q
=
Φ

22.32: a)
.
C
m
N
750
)
m
0.6
)(
C
N
125
(
2
2
⋅
=
=
=
Φ
EA

b) Since the field is parallel to the surface,
.0
=
Φ

c) Choose the Gaussian surface to equal the volume’s surface. Then: 750 –
EA=
,
C
N
577
)
750
C
10
40
.2
(
0
8

m
0.6
1
0
2
=
+
×
=
⇒
−
ε
E
ε
q
in the positive x -direction.

Since
0
<
q
we must have some net flux flowing in so

A
E
EA
−
→
on second face.

d)
0
<
q
but we have E pointing away from face I. This is due to an external field

that does not affect the flux but affects the value of E.

22.33: To find the charge enclosed, we need the flux through the parallelepiped:

C
m
N
5.
37
60

cos
)
C
N
10
50
.2
)(
m
0600
.0
)(
m
0500
.0
(
60
cos
2
4
1
1
⋅
=
°
×
=
°
=
Φ
AE

C
m
N
105
60
cos
)
C
N
10
00
.7
)(
m
0600
.0
)(
m
0500
.0
(
120
cos
2
4
2
2
⋅
−
=
°
×
=
°
=
Φ
AE

So the total flux is
,
C
m
N
5.
67
C
m
N
)
105
5.
37
(
2
2
2
1
⋅
−
=
⋅
−
=
Φ
+
Φ
=
Φ
and

.
C
10
97
.5
)
C
m
N
5.
67
(
10
0
2
0
−
×
−
=
⋅
−
=
Φ
=
ε
ε
q

b) There must be a net charge (negative) in the parallelepiped since there is a net

flux flowing into the surface. Also, there must be an external field or all lines would point
toward the slab.

22.34:

The α particle feels no force where the net electric field is zero. The fields can
cancel only in regions A and B.

sheet
line
E
E
=

0
0
2
2
λ
ε
σ
r
πε
=

cm
16
m
16
.0
)
C/m
100
(
C/m
50
λ
2
=
=
=
=
µ
π

µ
πσ
r

The fields cancel 16 cm from the line in regions A and B.

22.35:

The electric field
1
E
r

of the sheet of charge is toward the sheet, so the electric

field
2
E
r

of the sphere must be away from the sheet. This is true above the center of the
sphere. Let r be the distance above the center of the sphere for the point where the
electric field is zero.

3
2

0
0

1
2
1
4
1
2
so
R
r
Q
πε
ε
σ
E
E
=
=

m
097
.0
C
10
900
.0
)
m
120
.0
)(
C/m
10
00
.8
(
2
2

9

3
2
9

2

3
1
=
×
×
=
=
−

−
π
Q
R
πσ
r

22.36: a) For
,0
,
=
<
E
a
r
since no charge is enclosed.

For
,
,
2
0
4
1
r

q
πε
E
b
r
a
=
<
<
since there is +q inside a radius r.

For
=
<
<
E
c
r
b
,
0, since now the –q cancels the inner +q.

For
,
,
2
0
4
1
r

q
πε
E
c
r
=
>
since again the total charge enclosed is +q.

b)

c) Charge on inner shell surface is –q.
d) Charge on outer shell surface is +q.
e)

22.37: a)
,0
,
=
<
E
R
r
since no charge is enclosed.

b)
,
,
2
2
0
4
1
r

Q
πε
E
R
r
R
=
<
<
since charge enclosed is
,
,
2
.
2
0

2
4
1
r

Q
πε
E
R
r
Q
=
>
since

charge enclosed is 2Q.

22.38: a)
,
,
2
0
4
1
r

Q
πε
E
a
r
=
<
since the charge enclosed is Q.

,0
,
=
<
<
E
b
r
a
since the –Q on the inner surface of the shell cancels the +Q at the
center of the sphere.

2
0

2
4
1
,
r

Q
πε
E
b
r
−
=
>
, since the total enclosed charge is –2Q.

b) The surface charge density on inner surface:
2
4πa

Q
σ
−
=
.

c) The surface charge density on the outer surface:
.
2
4

2

πb

Q
σ
−
=

d)

e)